Chapter 13.10, Problem 57AAP

To determine

Current density and average current associated with the formation of single pit.

Answer to Problem 57AAP

The average current associated with the formation of single pit is $2.174\times {10}^{-8}\text{A}$ and the required current density is $9.50\times {10}^{-5}{\text{ A/cm}}^2$.

Explanation of Solution

Write the expression for the mass of the pit.

  $w={\rho }_{Al}{V}_{Al}$

Here, density of aluminum is ${\rho }_{Al}$ and volume of aluminum is $V_{Al}$.

Write the expression for corrosion rate.

  $\frac{w}{t}=\frac{IM}{nF}$

Rearrange the equation in terms of I.

  $I=\frac{wnF}{tM}$        (I)

Here, the corrosion rate is $w/t$, the current flow is $I$, atomic mass of the metal is $M$, number of electrons per atoms produced or consumed in the process is $n$, and the Faraday’s constant is $F$, $F=96500\text{C/mol}\text{or}96500\text{A}\cdot \text{s/mol}$.

The current density is expressed as,

  $i=\frac{I}{A}$        (II)

Conclusion:

It is assumed that the pit is in the shape of cylinder.

The density of the aluminum is $2.70{\text{g/cm}}^3$.

The mass of the cylindrical aluminum pit $\left(w\right)$ is expressed as follows.

 $w={\rho }_{\text{Al}}\times \text{Volume of the pit}$        (III)

Substitute $2.70{\text{g/cm}}^3$ for ${\rho }_{Al}$, 0.170 mm for $d$, and 1 mm for h in Equation (III).

 $\begin{array}{c}w=2.70{\text{g/cm}}^3\times \frac{\pi }{4}\times {\left(0.170\text{mm}\right)}^2\times 1\text{mm}\\ =2.70{\text{g/cm}}^3\times \frac{\pi }{4}\times {\left(0.170\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\right)}^2\times \left(1\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\right)\\ =2.70{\text{g/cm}}^3\times 2.2698\times {10}^{-5}{\text{cm}}^3\\ =6.1285\times {10}^{-5}\text{ g}\end{array}$

The top surface area of the cylindrical pit is expressed as follows.

  $\begin{array}{c}A=\frac{\pi }{4}{d}^2\\ =\frac{\pi }{4}\times {\left(0.170\text{mm}\right)}^2\\ =0.3848{\text{mm}}^2\times \frac{1{\text{cm}}^2}{100{\text{mm}}^2}\\ =2.298\times {10}^{-4}{\text{cm}}^2\end{array}$

Refer Table 13.1, “Standard electrode potentials at 25 °C”.

The anodic reaction of aluminum is as follows:

  ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$

Here, the number of electrons produced is, $n=3$.

Refer Figure 2.3, “Periodic Table of Elements”.

The atomic mass of aluminum is, $M_{\text{Al}}=26.98\text{g/mol}$.

Substitute $6.128\times {10}^{-5}\text{g}$ for $w$, 3 for $n$, $96500\text{A}\cdot \text{s/mol}$ for $F$, 350 days for $t$, and $26.98\text{g/mol}$ for $M$ in Equation (I).

  $\begin{array}{c}I=\frac{\left(6.128\times {10}^{-5}\text{g}\right)\left(3\right)\left(96500\text{A}\cdot \text{s/mol}\right)}{\left(350\text{day}\right)\left(24\frac{\text{h}}{\text{day}}\right)\left(3600\frac{\text{s}}{\text{h}}\right)\left(26.98\text{g/mol}\right)}\\ =2.174\times {10}^{-8}\text{A}\end{array}$

Thus, the average current associated with the formation of single pit is $2.174\times {10}^{-8}\text{A}$.

Substitute $2.174\times {10}^{-8}\text{A}$ for $I$ and $2.298\times {10}^{-4}{\text{cm}}^2$ for A in Equation (II).

  $\begin{array}{c}i=\frac{2.174\times {10}^{-8}\text{A}}{2.298\times {10}^{-4}{\text{cm}}^2}\\ =9.50\times {10}^{-5}{\text{ A/cm}}^2\end{array}$

Thus, required current density is $9.50\times {10}^{-5}{\text{ A/cm}}^2$.